[LeetCode] 783. Minimum Distance Between BST Nodes
Given a Binary Search Tree (BST) with the root node root
, return the minimum difference between the values of any two different nodes in the tree.
Example :
Input: root = [4,2,6,1,3,null,null] Output: 1 Explanation: Note that root is a TreeNode object, not an array. The given tree [4,2,6,1,3,null,null] is represented by the following diagram: 4 / \ 2 6 / \ 1 3 while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.
题意:给一个二叉搜索树,找到其中两个节点的最小差值
二叉搜索树,左孩子小于根节点,右孩子大于根节点
所以,二叉搜索树 有着中序有序的特点
我们可以对其中序遍历,把数提出来,判断相邻两节点
class Solution { private List<Integer> list; private void DFS(TreeNode root) { if (root.left != null) DFS(root.left); list.add(root.val); if (root.right != null) DFS(root.right); } public int minDiffInBST(TreeNode root) { list = new ArrayList<>(); int min = 3276800; DFS(root); for (int i = 0; i < list.size() - 1; i++) { int temp = list.get(i+1) - list.get(i); min = Math.min(temp, min); } return min; } }
当然这个算法显然有些多余,其一:额外开了O(n)的空间,其二多遍历了一遍
既然知道是中序有序的,那我们可以直接在遍历的时候就判断
class Solution { private int pre = -1; private int min = 3276800; public void DFS(TreeNode root) { if (root == null) return; DFS(root.left); if (pre != -1) min = Math.min(min, root.val - pre); pre = root.val; DFS(root.right); } public int minDiffInBST(TreeNode root) { DFS(root); return min; } }